Figure 1.1 These snowshoers on Mount Hood in Oregon are enjoying the heat flow and light caused by high
temperature. All three mechanisms of heat transfer are relevant to this picture. The heat flowing out of the fire also
turns the solid snow to liquid water and vapor. (credit: modification of work by “Mt. Hood Territory”/Flickr)
Chapter Outline
1.1 Temperature and Thermal Equilibrium
1.2 Thermometers and Temperature Scales
1.3 Thermal Expansion
1.4 Heat Transfer, Specific Heat, and Calorimetry
1.5 Phase Changes
1.6 Mechanisms of Heat Transfer
Heat and temperature are important concepts for each of us, every day. How we dress in the morning depends on
whether the day is hot or cold, and most of what we do requires energy that ultimately comes from the Sun. The
study of heat and temperature is part of an area of physics known as thermodynamics. The laws of thermodynamics
govern the flow of energy throughout the universe. They are studied in all areas of science and engineering, from
chemistry to biology to environmental science.
In this chapter, we explore heat and temperature. It is not always easy to distinguish these terms. Heat is the flow of
energy from one object to another. This flow of energy is caused by a difference in temperature. The transfer of heat
can change temperature, as can work, another kind of energy transfer that is central to thermodynamics. We return
to these basic ideas several times throughout the next four chapters, and you will see that they affect everything
from the behavior of atoms and molecules to cooking to our weather on Earth to the life cycles of stars.
Learning Objectives
By the end of this section, you will be able to:



Define temperature and describe it qualitatively
Explain thermal equilibrium
Explain the zeroth law of thermodynamics
Heat is familiar to all of us. We can feel heat entering our bodies from the summer Sun or from hot coffee or tea
after a winter stroll. We can also feel heat leaving our bodies as we feel the chill of night or the cooling effect of
sweat after exercise.
What is heat? How do we define it and how is it related to temperature? What are the effects of heat and how
does it flow from place to place? We will find that, in spite of the richness of the phenomena, a small set of
underlying physical principles unites these subjects and ties them to other fields. We start by examining
temperature and how to define and measure it.
Temperature
The concept of temperature has evolved from the common concepts of hot and cold. The scientific definition of
temperature explains more than our senses of hot and cold. As you may have already learned, many physical
quantities are defined solely in terms of how they are observed or measured, that is, they are
defined operationally. Temperature is operationally defined as the quantity of what we measure with a
thermometer. As we will see in detail in a later chapter on the kinetic theory of gases, temperature is
proportional to the average kinetic energy of translation, a fact that provides a more physical definition.
Differences in temperature maintain the transfer of heat, or heat transfer, throughout the universe. Heat
transfer is the movement of energy from one place or material to another as a result of a difference in
temperature. (You will learn more about heat transfer later in this chapter.)
Thermal Equilibrium
An important concept related to temperature is thermal equilibrium. Two objects are in thermal equilibrium if
they are in close contact that allows either to gain energy from the other, but nevertheless, no net energy is
transferred between them. Even when not in contact, they are in thermal equilibrium if, when they are placed in
contact, no net energy is transferred between them. If two objects remain in contact for a long time, they
typically come to equilibrium. In other words, two objects in thermal equilibrium do not exchange energy.
Experimentally, if object A is in equilibrium with object B, and object B is in equilibrium with object C, then (as
you may have already guessed) object A is in equilibrium with object C. That statement of transitivity is called
the zeroth law of thermodynamics. (The number “zeroth” was suggested by British physicist Ralph Fowler in
the 1930s. The first, second, and third laws of thermodynamics were already named and numbered then. The
zeroth law had seldom been stated, but it needs to be discussed before the others, so Fowler gave it a smaller
number.) Consider the case where A is a thermometer. The zeroth law tells us that if A reads a certain
temperature when in equilibrium with B, and it is then placed in contact with C, it will not exchange energy
with C; therefore, its temperature reading will remain the same (Figure 1.2). In other words, if two objects are in
thermal equilibrium, they have the same temperature.
Figure 1.2 If thermometer A is in thermal equilibrium with object B, and B is in thermal equilibrium with C, then A is in
thermal equilibrium with C. Therefore, the reading on A stays the same when A is moved over to make contact with C.
A thermometer measures its own temperature. It is through the concepts of thermal equilibrium and the zeroth
law of thermodynamics that we can say that a thermometer measures the temperature of something else, and to
make sense of the statement that two objects are at the same temperature.
In the rest of this chapter, we will often refer to “systems” instead of “objects.” As in the chapter on linear
momentum and collisions, a system consists of one or more objects—but in thermodynamics, we require a
system to be macroscopic, that is, to consist of a huge number (such as 10231023) of molecules. Then we can
say that a system is in thermal equilibrium with itself if all parts of it are at the same temperature. (We will
return to the definition of a thermodynamic system in the chapter on the first law of thermodynamics.)
Learning Objectives
By the end of this section, you will be able to:


Describe several different types of thermometers
Convert temperatures between the Celsius, Fahrenheit, and Kelvin scales
Any physical property that depends consistently and reproducibly on temperature can be used as the basis of a
thermometer. For example, volume increases with temperature for most substances. This property is the basis for
the common alcohol thermometer and the original mercury thermometers. Other properties used to measure
temperature include electrical resistance, color, and the emission of infrared radiation (Figure 1.3).
Figure 1.3 Because many physical properties depend on temperature, the variety of thermometers is remarkable.
(a) In this common type of thermometer, the alcohol, containing a red dye, expands more rapidly than the glass
encasing it. When the thermometer’s temperature increases, the liquid from the bulb is forced into the narrow tube,
producing a large change in the length of the column for a small change in temperature. (b) Each of the six squares
on this plastic (liquid crystal) thermometer contains a film of a different heat-sensitive liquid crystal material.
Below 95°F95°F, all six squares are black. When the plastic thermometer is exposed to a temperature of 95°F95°F,
the first liquid crystal square changes color. When the temperature reaches above 96.8°F96.8°F, the second liquid
crystal square also changes color, and so forth. (c) A firefighter uses a pyrometer to check the temperature of an
aircraft carrier’s ventilation system. The pyrometer measures infrared radiation (whose emission varies with
temperature) from the vent and quickly produces a temperature readout. Infrared thermometers are also frequently
used to measure body temperature by gently placing them in the ear canal. Such thermometers are more accurate
than the alcohol thermometers placed under the tongue or in the armpit. (credit b: modification of work by Tess
Watson; credit c: modification of work by Lamel J. Hinton, U.S. Navy)
Thermometers measure temperature according to well-defined scales of measurement. The three most common
temperature scales are Fahrenheit, Celsius, and Kelvin. Temperature scales are created by identifying two
reproducible temperatures. The freezing and boiling temperatures of water at standard atmospheric pressure are
commonly used.
On the Celsius scale, the freezing point of water is 0°C0°C and the boiling point is 100°C.100°C. The unit of
temperature on this scale is the degree Celsius (°C)(°C). The Fahrenheit scale (still the most frequently used for
common purposes in the United States) has the freezing point of water at 32°F32°F and the boiling point
at 212°F.212°F. Its unit is the degree Fahrenheit (°F°F). You can see that 100 Celsius degrees span the same
range as 180 Fahrenheit degrees. Thus, a temperature difference of one degree on the Celsius scale is 1.8 times as
large as a difference of one degree on the Fahrenheit scale, or ΔTF=95ΔTC.ΔTF=95ΔTC.
The definition of temperature in terms of molecular motion suggests that there should be a lowest possible
temperature, where the average kinetic energy of molecules is zero (or the minimum allowed by quantum
mechanics). Experiments confirm the existence of such a temperature, called absolute zero. An absolute
temperature scale is one whose zero point is absolute zero. Such scales are convenient in science because
several physical quantities, such as the volume of an ideal gas, are directly related to absolute temperature.
The Kelvin scale is the absolute temperature scale that is commonly used in science. The SI temperature unit is
the kelvin, which is abbreviated K (not accompanied by a degree sign). Thus 0 K is absolute zero. The freezing and
boiling points of water are 273.15 K and 373.15 K, respectively. Therefore, temperature differences are the same in
units of kelvins and degrees Celsius, or ΔTC=ΔTK.ΔTC=ΔTK.
The relationships between the three common temperature scales are shown in Figure 1.4. Temperatures on these
scales can be converted using the equations in Table 1.1.
Figure 1.4 Relationships between the Fahrenheit, Celsius, and Kelvin temperature scales are shown. The relative
sizes of the scales are also shown.
To convert from…
Use this equation…
Celsius to Fahrenheit
TF=95TC+32TF=95TC+32
Fahrenheit to Celsius
TC=59(TF−32)TC=59(TF−32)
Celsius to Kelvin
TK=TC+273.15TK=TC+273.15
Kelvin to Celsius
TC=TK−273.15TC=TK−273.15
Fahrenheit to Kelvin
TK=59(TF−32)+273.15TK=59(TF−32)+273.15
Kelvin to Fahrenheit
TF=95(TK−273.15)+32TF=95(TK−273.15)+32
Table 1.1 Temperature Conversions
To convert between Fahrenheit and Kelvin, convert to Celsius as an intermediate step.
EXAMPLE 1.1
Converting between Temperature Scales: Room Temperature
“Room temperature” is generally defined in physics to be 25°C25°C. (a) What is room temperature in °F°F? (b)
What is it in K?
Strategy
To answer these questions, all we need to do is choose the correct conversion equations and substitute the known
values.
Solution
To convert from °C°C to °F°F, use the equation
TF=95TC+32.TF=95TC+32.
Substitute the known value into the equation and solve:
TF=95(25°C)+32=77°F.TF=95(25°C)+32=77°F.
Similarly, we find that TK=TC+273.15=298KTK=TC+273.15=298K.
The Kelvin scale is part of the SI system of units, so its actual definition is more complicated than the one given
above. First, it is not defined in terms of the freezing and boiling points of water, but in terms of the triple point. The
triple point is the unique combination of temperature and pressure at which ice, liquid water, and water vapor can
coexist stably. As will be discussed in the section on phase changes, the coexistence is achieved by lowering the
pressure and consequently the boiling point to reach the freezing point. The triple-point temperature is defined as
273.16 K. This definition has the advantage that although the freezing temperature and boiling temperature of water
depend on pressure, there is only one triple-point temperature.
Second, even with two points on the scale defined, different thermometers give somewhat different results for other
temperatures. Therefore, a standard thermometer is required. Metrologists (experts in the science of measurement)
have chosen the constant-volume gas thermometer for this purpose. A vessel of constant volume filled with gas is
subjected to temperature changes, and the measured temperature is proportional to the change in pressure. Using
“TP” to represent the triple point,
T=ppTPTTP.T=ppTPTTP.
The results depend somewhat on the choice of gas, but the less dense the gas in the bulb, the better the results for
different gases agree. If the results are extrapolated to zero density, the results agree quite well, with zero pressure
corresponding to a temperature of absolute zero.
Constant-volume gas thermometers are big and come to equilibrium slowly, so they are used mostly as standards to
calibrate other thermometers.
INTERACTIVE
Visit this site to learn more about the constant-volume gas thermometer.
Learning Objectives
By the end of this section, you will be able to:


Answer qualitative questions about the effects of thermal expansion
Solve problems involving thermal expansion, including those involving thermal stress
The expansion of alcohol in a thermometer is one of many commonly encountered examples of thermal
expansion, which is the change in size or volume of a given system as its temperature changes. The most visible
example is the expansion of hot air. When air is heated, it expands and becomes less dense than the surrounding
air, which then exerts an (upward) force on the hot air and makes steam and smoke rise, hot air balloons float, and
so forth. The same behavior happens in all liquids and gases, driving natural heat transfer upward in homes,
oceans, and weather systems, as we will discuss in an upcoming section. Solids also undergo thermal expansion.
Railroad tracks and bridges, for example, have expansion joints to allow them to freely expand and contract with
temperature changes, as shown in Figure 1.5.
Figure 1.5 (a) Thermal expansion joints like these in the (b) Auckland Harbour Bridge in New Zealand allow bridges
to change length without buckling. (credit: modification of works by “ŠJů”/Wikimedia Commons)
What is the underlying cause of thermal expansion? As previously mentioned, an increase in temperature means an
increase in the kinetic energy of individual atoms. In a solid, unlike in a gas, the molecules are held in place by
forces from neighboring molecules; as we saw in Oscillations, the forces can be modeled as in harmonic springs
described by the Lennard-Jones potential. Energy in Simple Harmonic Motion shows that such potentials are
asymmetrical in that the potential energy increases more steeply when the molecules get closer to each other than
when they get farther away. Thus, at a given kinetic energy, the distance moved is greater when neighbors move
away from each other than when they move toward each other. The result is that increased kinetic energy
(increased temperature) increases the average distance between molecules—the substance expands.
For most substances under ordinary conditions, it is an excellent approximation that there is no preferred direction
(that is, the solid is “isotropic”), and an increase in temperature increases the solid’s size by a certain fraction in
each dimension. Therefore, if the solid is free to expand or contract, its proportions stay the same; only its overall
size changes.
LINEAR THERMAL EXPANSION
According to experiments, the dependence of thermal expansion on temperature, substance, and original initial
length is summarized in the equation
dLdT=αLdLdT=αL
1.1
where dLdTdLdT is the instantaneous change in length per temperature, L is the length, and αα is the coefficient
of linear expansion, a material property that varies slightly with temperature. As αα is nearly constant and also
very small, for practical purposes, we use the linear approximation:
ΔL=αLΔTΔL=αLΔT
1.2
where ΔLΔL is the change in length and ΔTΔT is the change in temperature.
Table 1.2 lists representative values of the coefficient of linear expansion. As noted earlier, ΔTΔT is the same
whether it is expressed in units of degrees Celsius or kelvins; thus, αα may have units of 1/°C1/°C or 1/K with the
same value in either case. Approximating αα as a constant is quite accurate for small changes in temperature and
sufficient for most practical purposes, even for large changes in temperature. We examine this approximation more
closely in the next example.
Material
Coefficient of Linear
Coefficient of Volume
Expansion α(1/°C)α(1/°C)
Expansion β(1/°C)β(1/°C)
Aluminum
25×10−625×10−6
75×10−675×10−6
Brass
19×10−619×10−6
56×10−656×10−6
Copper
17×10−617×10−6
51×10−651×10−6
Gold
14×10−614×10−6
42×10−642×10−6
Iron or steel
12×10−612×10−6
35×10−635×10−6
Invar (nickel-iron alloy)
0.9×10−60.9×10−6
2.7×10−62.7×10−6
Lead
29×10−629×10−6
87×10−687×10−6
Silver
18×10−618×10−6
54×10−654×10−6
Glass (ordinary)
9×10−69×10−6
27×10−627×10−6
Glass (Pyrex®)
3×10−63×10−6
9×10−69×10−6
0.4×10−60.4×10−6
1×10−61×10−6
~12×10−6~12×10−6
~36×10−6~36×10−6
2.5×10−62.5×10−6
7.5×10−67.5×10−6
Solids
Quartz
Concrete, brick
Marble (average)
Liquids
Ether
1650×10−61650×10−6
Ethyl alcohol
1100×10−61100×10−6
Gasoline
950×10−6950×10−6
Glycerin
500×10−6500×10−6
Mercury
180×10−6180×10−6
Water
210×10−6210×10−6
Gases
Material
Coefficient of Linear
Coefficient of Volume
Expansion α(1/°C)α(1/°C)
Expansion β(1/°C)β(1/°C)
Air and most other gases at
atmospheric pressure
3400×10−63400×10−6
Table 1.2 Thermal Expansion Coefficients
Thermal expansion is exploited in the bimetallic strip (Figure 1.6). This device can be used as a thermometer if the
curving strip is attached to a pointer on a scale. It can also be used to automatically close or open a switch at a
certain temperature, as in older or analog thermostats.
Figure 1.6 The curvature of a bimetallic strip depends on temperature. (a) The strip is straight at the starting
temperature, where its two components have the same length. (b) At a higher temperature, this strip bends to the
right, because the metal on the left has expanded more than the metal on the right. At a lower temperature, the strip
would bend to the left.
EXAMPLE 1.2
Calculating Linear Thermal Expansion
The main span of San Francisco’s Golden Gate Bridge is 1275 m long at its coldest. The bridge is exposed to
temperatures ranging from –15°C–15°C to 40°C40°C. What is its change in length between these temperatures?
Assume that the bridge is made entirely of steel.
Strategy
Use the equation for linear thermal expansion ΔL=αLΔTΔL=αLΔT to calculate the change in length, ΔLΔL. Use
the coefficient of linear expansion αα for steel from Table 1.2, and note that the change in
temperature ΔTΔT is 55°C.55°C.
Solution
Substitute all of the known values into the equation to solve for ΔLΔL:
ΔL=αLΔT=(12×10−6°C)(1275m)(55°C)=0.84m.ΔL=αLΔT=(12×10−6°C)(1275m)(55°C)=0.84m.
Significance
Although not large compared with the length of the bridge, this change in length is observable. It is generally spread
over many expansion joints so that the expansion at each joint is small.
Thermal Expansion in Two and Three Dimensions
Unconstrained objects expand in all dimensions, as illustrated in Figure 1.7. That is, their areas and volumes, as
well as their lengths, increase with temperature. Because the proportions stay the same, holes and container
volumes also get larger with temperature. If you cut a hole in a metal plate, the remaining material will expand
exactly as it would if the piece you removed were still in place. The piece would get bigger, so the hole must get
bigger too.
THERMAL EXPANSION IN TWO DIMENSIONS
For small temperature changes, the change in area ΔAΔA is given by
ΔA=2αAΔTΔA=2αAΔT
1.3
where ΔAΔA is the change in area A,ΔTA,ΔT is the change in temperature, and αα is the coefficient of linear
expansion, which varies slightly with temperature. (The derivation of this equation is analogous to that of the more
important equation for three dimensions, below.)
Figure 1.7 In general, objects expand in all directions as temperature increases. In these drawings, the original
boundaries of the objects are shown with solid lines, and the expanded boundaries with dashed lines. (a) Area
increases because both length and width increase. The area of a circular plug also increases. (b) If the plug is
removed, the hole it leaves becomes larger with increasing temperature, just as if the expanding plug were still in
place. (c) Volume also increases, because all three dimensions increase.
THERMAL EXPANSION IN THREE DIMENSIONS
The relationship between volume and temperature dVdTdVdT is given by dVdT=βVdVdT=βV, where ββ is
the coefficient of volume expansion. As you can show in Exercise 1.60, β=3αβ=3α. This equation is usually
written as
ΔV=βVΔT.ΔV=βVΔT.
1.4
Note that the values of ββ in Table 1.2 are equal to 3α3α except for rounding.
Volume expansion is defined for liquids, but linear and area expansion are not, as a liquid’s changes in linear
dimensions and area depend on the shape of its container. Thus, Table 1.2 shows liquids’ values of ββ but not αα.
In general, objects expand with increasing temperature. Water is the most important exception to this rule. Water
does expand with increasing temperature (its density decreases) at temperatures greater
than 4°C(40°F)4°C(40°F). However, it is densest at +4°C+4°C and expands with decreasing temperature
between +4°C+4°C and 0°C0°C (40°Fto32°F40°Fto32°F), as shown in Figure 1.8. A striking effect of this
phenomenon is the freezing of water in a pond. When water near the surface cools down to 4°C,4°C, it is denser
than the remaining water and thus sinks to the bottom. This “turnover” leaves a layer of warmer water near the
surface, which is then cooled. However, if the temperature in the surface layer drops below 4°C4°C, that water is
less dense than the water below, and thus stays near the top. As a result, the pond surface can freeze over. The
layer of ice insulates the liquid water below it from low air temperatures. Fish and other aquatic life can survive
in 4°C4°C water beneath ice, due to this unusual characteristic of water.
Figure 1.8 This curve shows the density of water as a function of temperature. Note that the thermal expansion at
low temperatures is very small. The maximum density at 4°C4°C is only 0.0075%0.0075% greater than the density
at 2°C2°C, and 0.012%0.012% greater than that at 0°C0°C. The decrease of density below 4°C4°C occurs because the
liquid water approachs the solid crystal form of ice, which contains more empty space than the liquid.
EXAMPLE 1.3
Calculating Thermal Expansion
Suppose your 60.0-L (15.9-gal(15.9-gal-gal) steel gasoline tank is full of gas that is cool because it has just been
pumped from an underground reservoir. Now, both the tank and the gasoline have a temperature
of 15.0°C.15.0°C. How much gasoline has spilled by the time they warm to 35.0°C35.0°C?
Strategy
The tank and gasoline increase in volume, but the gasoline increases more, so the amount spilled is the difference
in their volume changes. We can use the equation for volume expansion to calculate the change in volume of the
gasoline and of the tank. (The gasoline tank can be treated as solid steel.)
Solution
1. Use the equation for volume expansion to calculate the increase in volume of the steel tank:
ΔVs=βsVsΔT.ΔVs=βsVsΔT.
2. The increase in volume of the gasoline is given by this equation:
ΔVgas=βgasVgasΔT.ΔVgas=βgasVgasΔT.
3. Find the difference in volume to determine the amount spilled as
Vspill=ΔVgas−ΔVs.Vspill=ΔVgas−ΔVs.
Alternatively, we can combine these three equations into a single equation. (Note that the original volumes are
equal.)
Vspill=(βgas−βs)VΔT=[(950−35)×10−6/°C](60.0L)(20.0°C)=1.10L.Vspill=(βgas−βs)VΔT=[(950−3
5)×10−6/°C](60.0L)(20.0°C)=1.10L.
Significance
This amount is significant, particularly for a 60.0-L tank. The effect is so striking because the gasoline and steel
expand quickly. The rate of change in thermal properties is discussed later in this chapter.
If you try to cap the tank tightly to prevent overflow, you will find that it leaks anyway, either around the cap or by
bursting the tank. Tightly constricting the expanding gas is equivalent to compressing it, and both liquids and solids
resist compression with extremely large forces. To avoid rupturing rigid containers, these containers have air gaps,
which allow them to expand and contract without stressing them.
CHECK YOUR UNDERSTANDING 1.1
Check Your Understanding Does a given reading on a gasoline gauge indicate more gasoline in cold weather or in
hot weather, or does the temperature not matter?
Thermal Stress
If you change the temperature of an object while preventing it from expanding or contracting, the object is subjected
to stress that is compressive if the object would expand in the absence of constraint and tensile if it would contract.
This stress resulting from temperature changes is known as thermal stress. It can be quite large and can cause
damage.
To avoid this stress, engineers may design components so they can expand and contract freely. For instance, in
highways, gaps are deliberately left between blocks to prevent thermal stress from developing. When no gaps can
be left, engineers must consider thermal stress in their designs. Thus, the reinforcing rods in concrete are made of
steel because steel’s coefficient of linear expansion is nearly equal to that of concrete.
To calculate the thermal stress in a rod whose ends are both fixed rigidly, we can think of the stress as developing in
two steps. First, let the ends be free to expand (or contract) and find the expansion (or contraction). Second, find the
stress necessary to compress (or extend) the rod to its original length by the methods you studied in Static
Equilibrium and Elasticity on static equilibrium and elasticity. In other words, the ΔLΔL of the thermal expansion
equals the ΔLΔL of the elastic distortion (except that the signs are opposite).
EXAMPLE 1.4
Calculating Thermal Stress
Concrete blocks are laid out next to each other on a highway without any space between them, so they cannot
expand. The construction crew did the work on a winter day when the temperature was 5°C5°C. Find the stress in
the blocks on a hot summer day when the temperature is 38°C38°C. The compressive Young’s modulus of
concrete is Y=20×109N/m2Y=20×109N/m2.
Strategy
According to the chapter on static equilibrium and elasticity, the stress F/A is given by
FA=YΔLL0,FA=YΔLL0,
where Y is the Young’s modulus of the material—concrete, in this case. In thermal
expansion, ΔL=αL0ΔT.ΔL=αL0ΔT. We combine these two equations by noting that the two ΔL’sΔL’s are
equal, as stated above. Because we are not given L0L0 or A, we can obtain a numerical answer only if they both
cancel out.
Solution
We substitute the thermal-expansion equation into the elasticity equation to get
FA=YαL0ΔTL0=YαΔT,FA=YαL0ΔTL0=YαΔT,
and as we hoped, L0L0 has canceled and A appears only in F/A, the notation for the quantity we are calculating.
Now we need only insert the numbers:
FA=(20×109N/m2)(12×10−6/°C)(38°C−5°C)=7.9×106N/m2.FA=(20×109N/m2)(12×10−6/°C)(38
°C−5°C)=7.9×106N/m2.
Significance
The ultimate compressive strength of concrete is 20×106N/m2,20×106N/m2, so the blocks are unlikely to
break. However, the ultimate shear strength of concrete is only 2×106N/m2,2×106N/m2, so some might chip
off.
CHECK YOUR UNDERSTANDING 1.2
Check Your Understanding Two objects A and B have the same dimensions and are constrained identically. A is
made of a material with a higher thermal expansion coefficient than B. If the objects are heated identically, will A feel
a greater stress than B?
Learning Objectives
By the end of this section, you will be able to:


Explain phenomena involving heat as a form of energy transfer
Solve problems involving heat transfer
We have seen in previous chapters that energy is one of the fundamental concepts of physics. Heat is a type of
energy transfer that is caused by a temperature difference, and it can change the temperature of an object. As we
learned earlier in this chapter, heat transfer is the movement of energy from one place or material to another as a
result of a difference in temperature. Heat transfer is fundamental to such everyday activities as home heating and
cooking, as well as many industrial processes. It also forms a basis for the topics in the remainder of this chapter.
We also introduce the concept of internal energy, which can be increased or decreased by heat transfer. We
discuss another way to change the internal energy of a system, namely doing work on it. Thus, we are beginning the
study of the relationship of heat and work, which is the basis of engines and refrigerators and the central topic (and
origin of the name) of thermodynamics.
Internal Energy and Heat
A thermal system has internal energy (also called thermal energy), which is the sum of the mechanical energies of
its molecules. A system’s internal energy is proportional to its temperature. As we saw earlier in this chapter, if two
objects at different temperatures are brought into contact with each other, energy is transferred from the hotter to
the colder object until the bodies reach thermal equilibrium (that is, they are at the same temperature). No work is
done by either object because no force acts through a distance (as we discussed in Work and Kinetic Energy).
These observations reveal that heat is energy transferred spontaneously due to a temperature difference. Figure
1.9 shows an example of heat transfer.
Figure 1.9 (a) Here, the soft drink has a higher temperature than the ice, so they are not in thermal equilibrium. (b)
When the soft drink and ice are allowed to interact, heat is transferred from the drink to the ice due to the difference
in temperatures until they reach the same temperature, T'T′, achieving equilibrium. In fact, since the soft drink and
ice are both in contact with the surrounding air and the bench, the ultimate equilibrium temperature will be the same
as that of the surroundings.
The meaning of “heat” in physics is different from its ordinary meaning. For example, in conversation, we may say
“the heat was unbearable,” but in physics, we would say that the temperature was high. Heat is a form of energy
flow, whereas temperature is not. Incidentally, humans are sensitive to heat flow rather than to temperature.
Since heat is a form of energy, its SI unit is the joule (J). Another common unit of energy often used for heat is
the calorie (cal), defined as the energy needed to change the temperature of 1.00 g of water by 1.00°C1.00°C —
specifically, between 14.5°C14.5°C and 15.5°C15.5°C, since there is a slight temperature dependence. Also
commonly used is the kilocalorie (kcal), which is the energy needed to change the temperature of 1.00 kg of water
by 1.00°C1.00°C. Since mass is most often specified in kilograms, the kilocalorie is convenient. Confusingly, food
calories (sometimes called “big calories,” abbreviated Cal) are actually kilocalories, a fact not easily determined from
package labeling.
Mechanical Equivalent of Heat
It is also possible to change the temperature of a substance by doing work, which transfers energy into or out of a
system. This realization helped establish that heat is a form of energy. James Prescott Joule (1818–1889)
performed many experiments to establish the mechanical equivalent of heat—the work needed to produce the
same effects as heat transfer. In the units used for these two quantities, the value for this equivalence is
1.000kcal=4186J.1.000kcal=4186J.
We consider this equation to represent the conversion between two units of energy. (Other numbers that you may
see refer to calories defined for temperature ranges other than 14.5°C14.5°C to 15.5°C15.5°C.)
Figure 1.10 shows one of Joule’s most famous experimental setups for demonstrating that work and heat can
produce the same effects and measuring the mechanical equivalent of heat. It helped establish the principle of
conservation of energy. Gravitational potential energy (U) was converted into kinetic energy (K), and then
randomized by viscosity and turbulence into increased average kinetic energy of atoms and molecules in the
system, producing a temperature increase. Joule’s contributions to thermodynamics were so significant that the SI
unit of energy was named after him.
Figure 1.10 Joule’s experiment established the equivalence of heat and work. As the masses descended, they
caused the paddles to do work, W=mghW=mgh, on the water. The result was a temperature increase, ΔTΔT,
measured by the thermometer. Joule found that ΔTΔT was proportional to W and thus determined the mechanical
equivalent of heat.
Increasing internal energy by heat transfer gives the same result as increasing it by doing work. Therefore, although
a system has a well-defined internal energy, we cannot say that it has a certain “heat content” or “work content.” A
well-defined quantity that depends only on the current state of the system, rather than on the history of that system,
is known as a state variable. Temperature and internal energy are state variables. To sum up this paragraph, heat
and work are not state variables.
Incidentally, increasing the internal energy of a system does not necessarily increase its temperature. As we’ll see in
the next section, the temperature does not change when a substance changes from one phase to another. An
example is the melting of ice, which can be accomplished by adding heat or by doing frictional work, as when an ice
cube is rubbed against a rough surface.
Temperature Change and Heat Capacity
We have noted that heat transfer often causes temperature change. Experiments show that with no phase change
and no work done on or by the system, the transferred heat is typically directly proportional to the change in
temperature and to the mass of the system, to a good approximation. (Below we show how to handle situations
where the approximation is not valid.) The constant of proportionality depends on the substance and its phase,
which may be gas, liquid, or solid. We omit discussion of the fourth phase, plasma, because although it is the most
common phase in the universe, it is rare and short-lived on Earth.
We can understand the experimental facts by noting that the transferred heat is the change in the internal energy,
which is the total energy of the molecules. Under typical conditions, the total kinetic energy of the
molecules KtotalKtotal is a constant fraction of the internal energy (for reasons and with exceptions that we’ll see in
the next chapter). The average kinetic energy of a molecule KaveKave is proportional to the absolute temperature.
Therefore, the change in internal energy of a system is typically proportional to the change in temperature and to the
number of molecules, N. Mathematically, ΔU∝ΔKtotal=NKave∝NΔTΔU∝ΔKtotal=NKave∝NΔT The
dependence on the substance results in large part from the different masses of atoms and molecules. We are
considering its heat capacity in terms of its mass, but as we will see in the next chapter, in some cases, heat
capacities per molecule are similar for different substances. The dependence on substance and phase also results
from differences in the potential energy associated with interactions between atoms and molecules.
HEAT TRANSFER AND TEMPERATURE CHANGE
A practical approximation for the relationship between heat transfer and temperature change is:
Q=mcΔT,Q=mcΔT,
1.5
where Q is the symbol for heat transfer (“quantity of heat”), m is the mass of the substance, and ΔTΔT is the
change in temperature. The symbol c stands for the specific heat (also called “specific heat capacity”) and depends
on the material and phase. The specific heat is numerically equal to the amount of heat necessary to change the
temperature of 1.001.00 kg of mass by 1.00°C1.00°C. The SI unit for specific heat
is J/(kg×K)J/(kg×K) or J/(kg×°C)J/(kg×°C). (Recall that the temperature change ΔTΔT is the same in units
of kelvin and degrees Celsius.)
Values of specific heat must generally be measured, because there is no simple way to calculate them
precisely. Table 1.3 lists representative values of specific heat for various substances. We see from this table that
the specific heat of water is five times that of glass and 10 times that of iron, which means that it takes five times as
much heat to raise the temperature of water a given amount as for glass, and 10 times as much as for iron. In fact,
water has one of the largest specific heats of any material, which is important for sustaining life on Earth.
The specific heats of gases depend on what is maintained constant during the heating—typically either the volume
or the pressure. In the table, the first specific heat value for each gas is measured at constant volume, and the
second (in parentheses) is measured at constant pressure. We will return to this topic in the chapter on the kinetic
theory of gases.
Substances
Solids
Specific Heat (c)
J/(kg⋅°C)J/(kg·°C)
kcal/(kg⋅°C)[2]kcal/(kg·°C)[2]
Aluminum
900
0.215
Asbestos
800
0.19
Concrete, granite (average)
840
0.20
Copper
387
0.0924
Substances
Specific Heat (c)
Glass
840
0.20
Gold
129
0.0308
Human body (average at 37°C37°C)
3500
0.83
Ice (average, −50°Cto0°C−50°Cto0°C)
2090
0.50
Iron, steel
452
0.108
Lead
128
0.0305
Silver
235
0.0562
Wood
1700
0.40
Benzene
1740
0.415
Ethanol
2450
0.586
Glycerin
2410
0.576
Mercury
139
0.0333
4186
1.000
Liquids
Water (15.0°C)(15.0°C)
Gases[3]
Air (dry)
721 (1015)
0.172 (0.242)
1670 (2190)
0.399 (0.523)
638 (833)
0.152 (0.199)
Nitrogen
739 (1040)
0.177 (0.248)
Oxygen
651 (913)
0.156 (0.218)
1520 (2020)
0.363 (0.482)
Ammonia
Carbon dioxide
Steam (100°C)(100°C)
Table 1.3 Specific Heats of Various Substances[1] [1]The values for solids and liquids are at constant volume and 25°C25°C,
except as noted. [2]These values are identical in units of
cal/g⋅°C.cal/g·°C.
[3]
Specific heats at constant volume and
at 20.0°C20.0°C except as noted, and at 1.00 atm pressure. Values in parentheses are specific heats at a constant pressure
of 1.00 atm.
In general, specific heat also depends on temperature. Thus, a precise definition of c for a substance must be given
in terms of an infinitesimal change in temperature. To do this, we note that c=1mΔQΔTc=1mΔQΔT and
replace ΔΔ with d:
c=1mdQdT.c=1mdQdT.
Except for gases, the temperature and volume dependence of the specific heat of most substances is weak at
normal temperatures. Therefore, we will generally take specific heats to be constant at the values given in the table.
EXAMPLE 1.5
Calculating the Required Heat
A 0.500-kg aluminum pan on a stove and 0.250 L of water in it are heated from 20.0°C20.0°C to 80.0°C80.0°C.
(a) How much heat is required? What percentage of the heat is used to raise the temperature of (b) the pan and (c)
the water?
Strategy
We can assume that the pan and the water are always at the same temperature. When you put the pan on the
stove, the temperature of the water and that of the pan are increased by the same amount. We use the equation for
the heat transfer for the given temperature change and mass of water and aluminum. The specific heat values for
water and aluminum are given in Table 1.3.
Solution
1. Calculate the temperature difference:
ΔT=Tf−Ti=60.0°C.ΔT=Tf−Ti=60.0°C.
2. Calculate the mass of water. Because the density of water is 1000kg/m31000kg/m3, 1 L of water has a
mass of 1 kg, and the mass of 0.250 L of water is mw=0.250kgmw=0.250kg.
3. Calculate the heat transferred to the water. Use the specific heat of water in Table 1.3:
Qw=mwcwΔT=(0.250kg)(4186J/kg°C)(60.0°C)=62.8kJ.Qw=mwcwΔT=(0.250kg)(4186J/kg°
C)(60.0°C)=62.8kJ.
4. Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in Table 1.3:
QAl=mA1cA1ΔT=(0.500kg)(900J/kg°C)(60.0°C)=27.0kJ.QAl=mA1cA1ΔT=(0.500kg)(900J/k
g°C)(60.0°C)=27.0kJ.
5. Find the total transferred heat:
QTotal=QW+QAl=89.8kJ.QTotal=QW+QAl=89.8kJ.
Significance
In this example, the heat transferred to the water is more than the aluminum pan. Although the mass of the pan is
twice that of the water, the specific heat of water is over four times that of aluminum. Therefore, it takes a bit more
than twice as much heat to achieve the given temperature change for the water as for the aluminum pan.
Example 1.6 illustrates a temperature rise caused by doing work. (The result is the same as if the same amount of
energy had been added with a blowtorch instead of mechanically.)
EXAMPLE 1.6
Calculating the Temperature Increase from the Work Done on a Substance
Truck brakes used to control speed on a downhill run do work, converting gravitational potential energy into
increased internal energy (higher temperature) of the brake material (Figure 1.11). This conversion prevents the
gravitational potential energy from being converted into kinetic energy of the truck. Since the mass of the truck is
much greater than that of the brake material absorbing the energy, the temperature increase may occur too fast for
sufficient heat to transfer from the brakes to the environment; in other words, the brakes may overheat.
Figure 1.11 The smoking brakes on a braking truck are visible evidence of the mechanical equivalent of heat.
Calculate the temperature increase of 10 kg of brake material with an average specific heat
of 800J/kg⋅°C800J/kg·°C if the material retains 10% of the energy from a 10,000-kg truck descending 75.0 m
(in vertical displacement) at a constant speed.
Strategy
We calculate the gravitational potential energy (Mgh) that the entire truck loses in its descent, equate it to the
increase in the brakes’ internal energy, and then find the temperature increase produced in the brake material alone.
Solution
First we calculate the change in gravitational potential energy as the truck goes downhill:
Mgh=(10,000kg)(9.80m/s2)(75.0m)=7.35×106J.Mgh=(10,000kg)(9.80m/s2)(75.0m)=7.35×106J.
Because the kinetic energy of the truck does not change, conservation of energy tells us the lost potential energy is
dissipated, and we assume that 10% of it is transferred to internal energy of the brakes, so
take Q=Mgh/10Q=Mgh/10. Then we calculate the temperature change from the heat transferred, using
ΔT=Qmc,ΔT=Qmc,
where m is the mass of the brake material. Insert the given values to find
ΔT=7.35×105J(10kg)(800J/kg°C)=92°C.ΔT=7.35×105J(10kg)(800J/kg°C)=92°C.
Significance
If the truck had been traveling for some time, then just before the descent, the brake temperature would probably be
higher than the ambient temperature. The temperature increase in the descent would likely raise the temperature of
the brake material very high, so this technique is not practical. Instead, the truck would use the technique of engine
braking. A different idea underlies the recent technology of hybrid and electric cars, where mechanical energy
(kinetic and gravitational potential energy) is converted by the brakes into electrical energy in the battery, a process
called regenerative braking.
In a common kind of problem, objects at different temperatures are placed in contact with each other but isolated
from everything else, and they are allowed to come into equilibrium. A container that prevents heat transfer in or out
is called a calorimeter, and the use of a calorimeter to make measurements (typically of heat or specific heat
capacity) is called calorimetry.
We will use the term “calorimetry problem” to refer to any problem in which the objects concerned are thermally
isolated from their surroundings. An important idea in solving calorimetry problems is that during a heat transfer
between objects isolated from their surroundings, the heat gained by the colder object must equal the heat lost by
the hotter object, due to conservation of energy:
Qcold+Qhot=0.Qcold+Qhot=0.
1.6
We express this idea by writing that the sum of the heats equals zero because the heat gained is usually considered
positive; the heat lost, negative.
EXAMPLE 1.7
Calculating the Final Temperature in Calorimetry
Suppose you pour 0.250 kg of 20.0-°C20.0-°C water (about a cup) into a 0.500-kg aluminum pan off the stove
with a temperature of 150°C150°C. Assume no heat transfer takes place to anything else: The pan is placed on
an insulated pad, and heat transfer to the air is neglected in the short time needed to reach equilibrium. Thus, this is
a calorimetry problem, even though no isolating container is specified. Also assume that a negligible amount of
water boils off. What is the temperature when the water and pan reach thermal equilibrium?
Strategy
Originally, the pan and water are not in thermal equilibrium: The pan is at a higher temperature than the water. Heat
transfer restores thermal equilibrium once the water and pan are in contact; it stops once thermal equilibrium
between the pan and the water is achieved. The heat lost by the pan is equal to the heat gained by the water—that
is the basic principle of calorimetry.
Solution
1. Use the equation for heat transfer Q=mcΔTQ=mcΔT to express the heat lost by the aluminum pan in
terms of the mass of the pan, the specific heat of aluminum, the initial temperature of the pan, and the final
temperature:
Qhot=mA1cA1(Tf−150°C).Qhot=mA1cA1(Tf−150°C).
2. Express the heat gained by the water in terms of the mass of the water, the specific heat of water, the initial
temperature of the water, and the final temperature:
Qcold=mwcw(Tf−20.0°C).Qcold=mwcw(Tf−20.0°C).
3. Note that Qhot<0Qhot<0 and Qcold>0Qcold>0 and that as stated above, they must sum to zero:
Qcold+QhotQcoldmwcw(Tf−20.0°C)===0−Qhot−mA1cA1(Tf−150°C).Qcold+Qhot=0Qcold=−Q
hotmwcw(Tf−20.0°C)=−mA1cA1(Tf−150°C).
4. Bring all terms involving TfTf on the left hand side and all other terms on the right hand side. Solving
for Tf,Tf,
Tf=mA1cA1(150°C)+mwcw(20.0°C)mA1cA1+mwcw,Tf=mA1cA1(150°C)+mwcw(20.0°C)mA1cA
1+mwcw,
and insert the numerical values:
Tf=(0.500kg)(900J/kg°C)(150°C)+(0.250kg)(4186J/kg°C)(20.0°C)(0.500kg)(900J
/kg°C)+(0.250kg)(4186J/kg°C)=59.1°C.Tf=(0.500kg)(900J/kg°C)(150°C)+(0.250kg)(4186J/k
g°C)(20.0°C)(0.500kg)(900J/kg°C)+(0.250kg)(4186J/kg°C)=59.1°C.
Significance
Why is the final temperature so much closer to 20.0°C20.0°C than to 150°C150°C? The reason is that water
has a greater specific heat than most common substances and thus undergoes a smaller temperature change for a
given heat transfer. A large body of water, such as a lake, requires a large amount of heat to increase its
temperature appreciably. This explains why the temperature of a lake stays relatively constant during the day even
when the temperature change of the air is large. However, the water temperature does change over longer times
(e.g., summer to winter).
CHECK YOUR UNDERSTANDING 1.3
If 25 kJ is necessary to raise the temperature of a rock from 25°Cto30°C,25°Cto30°C, how much heat is
necessary to heat the rock from 45°Cto50°C45°Cto50°C?
EXAMPLE 1.8
Temperature-Dependent Heat Capacity
At low temperatures, the specific heats of solids are typically proportional to T3T3. The first understanding of this
behavior was due to the Dutch physicist Peter Debye, who in 1912, treated atomic oscillations with the quantum
theory that Max Planck had recently used for radiation. For instance, a good approximation for the specific heat of
salt, NaCl, is c=3.33×104Jkg⋅K(T321K)3.c=3.33×104Jkg·K(T321K)3. The constant 321 K is called the Debye
temperature of NaCl, ΘD,ΘD, and the formula works well when T<0.04ΘD.T<0.04ΘD. Using this formula, how
much heat is required to raise the temperature of 24.0 g of NaCl from 5 K to 15 K?
Solution
Because the heat capacity depends on the temperature, we need to use the equation
c=1mdQdT.c=1mdQdT.
We solve this equation for Q by integrating both sides: Q=m∫T2T1cdT.Q=m∫T1T2cdT.
Then we substitute the given values in and evaluate the integral:
Q=(0.024kg)∫T2T13.33×10–
6Jkg⋅K(T321K)3dT=(6.04×10−4JK4)T4∣∣∣15K5K=0.302J.Q=(0.024kg)∫T1T23.33×10–
6Jkg·K(T321K)3dT=(6.04×10−4JK4)T4|5K15K=0.302J.
Significance
If we had used the equation Q=mcΔTQ=mcΔT and the room-temperature specific heat of
salt, 880J/kg⋅K,880J/kg·K, we would have gotten a very different value.
Learning Objectives
By the end of this section, you will be able to:



Describe phase transitions and equilibrium between phases
Solve problems involving latent heat
Solve calorimetry problems involving phase changes
Phase transitions play an important theoretical and practical role in the study of heat flow. In melting (or “fusion”), a
solid turns into a liquid; the opposite process is freezing. In evaporation, a liquid turns into a gas; the opposite
process is condensation.
A substance melts or freezes at a temperature called its melting point, and boils (evaporates rapidly) or condenses
at its boiling point. These temperatures depend on pressure. High pressure favors the denser form, so typically, high
pressure raises the melting point and boiling point, and low pressure lowers them. For example, the boiling point of
water is 100°C100°C at 1.00 atm. At higher pressure, the boiling point is higher, and at lower pressure, it is lower.
The main exception is the melting and freezing of water, discussed in the next section.
Phase Diagrams
The phase of a given substance depends on the pressure and temperature. Thus, plots of pressure versus
temperature showing the phase in each region provide considerable insight into thermal properties of substances.
Such a pT graph is called a phase diagram.
Figure 1.12 shows the phase diagram for water. Using the graph, if you know the pressure and temperature, you
can determine the phase of water. The solid curves—boundaries between phases—indicate phase transitions, that
is, temperatures and pressures at which the phases coexist. For example, the boiling point of water
is 100°C100°C at 1.00 atm. As the pressure increases, the boiling temperature rises gradually
to 374°C374°C at a pressure of 218 atm. A pressure cooker (or even a covered pot) cooks food faster than an
open pot, because the water can exist as a liquid at temperatures greater than 100°C100°C without all boiling
away. (As we’ll see in the next section, liquid water conducts heat better than steam or hot air.) The boiling point
curve ends at a certain point called the critical point—that is, a critical temperature, above which the liquid and
gas phases cannot be distinguished; the substance is called a supercritical fluid. At sufficiently high pressure above
the critical point, the gas has the density of a liquid but does not condense. Carbon dioxide, for example, is
supercritical at all temperatures above 31.0°C31.0°C. Critical pressure is the pressure of the critical point.
Figure 1.12 The phase diagram (pT graph) for water shows solid (s), liquid (l), and vapor (v) phases. At
temperatures and pressure above those of the critical point, there is no distinction between liquid and vapor. Note
that the axes are nonlinear and the graph is not to scale. This graph is simplified—it omits several exotic phases of
ice at higher pressures. The phase diagram of water is unusual because the melting-point curve has a negative
slope, showing that you can melt ice by increasing the pressure.
Similarly, the curve between the solid and liquid regions in Figure 1.12 gives the melting temperature at various
pressures. For example, the melting point is 0°C0°C at 1.00 atm, as expected. Water has the unusual property that
ice is less dense than liquid water at the melting point, so at a fixed temperature, you can change the phase from
solid (ice) to liquid (water) by increasing the pressure. That is, the melting temperature of ice falls with increased
pressure, as the phase diagram shows. For example, when a car is driven over snow, the increased pressure from
the tires melts the snowflakes; afterwards, the water refreezes and forms an ice layer.
As you learned in the earlier section on thermometers and temperature scales, the triple point is the combination of
temperature and pressure at which ice, liquid water, and water vapor can coexist stably—that is, all three phases
exist in equilibrium. For water, the triple point occurs at 273.16K(0.01°C)273.16K(0.01°C) and 611.2 Pa; that
is a more accurate calibration temperature than the melting point of water at 1.00 atm,
or 273.15K(0.0°C)273.15K(0.0°C).
INTERACTIVE
View this video to see a substance at its triple point.
At pressures below that of the triple point, there is no liquid phase; the substance can exist as either gas or solid.
For water, there is no liquid phase at pressures below 0.00600 atm. The phase change from solid to gas is
called sublimation. You may have noticed that snow can disappear into thin air without a trace of liquid water, or
that ice cubes can disappear in a freezer. Both are examples of sublimation. The reverse also happens: Frost can
form on very cold windows without going through the liquid stage. Figure 1.13 shows the result, as well as showing
a familiar example of sublimation. Carbon dioxide has no liquid phase at atmospheric pressure. Solid CO2CO2 is
known as dry ice because instead of melting, it sublimes. Its sublimation temperature at atmospheric pressure
is −78°C−78°C. Certain air fresheners use the sublimation of a solid to spread a perfume around a room. Some
solids, such as osmium tetroxide, are so toxic that they must be kept in sealed containers to prevent human
exposure to their sublimation-produced vapors.
Figure 1.13 Direct transitions between solid and vapor are common, sometimes useful, and even beautiful. (a) Dry
ice sublimes directly to carbon dioxide gas. The visible “smoke” consists of water droplets that condensed in the air
cooled by the dry ice. (b) Frost forms patterns on a very cold window, an example of a solid formed directly from a
vapor. (credit a: modification of work by Windell Oskay; credit b: modification of work by Liz West)
Equilibrium
At the melting temperature, the solid and liquid phases are in equilibrium. If heat is added, some of the solid will
melt, and if heat is removed, some of the liquid will freeze. The situation is somewhat more complex for liquid-gas
equilibrium. Generally, liquid and gas are in equilibrium at any temperature. We call the gas phase a vapor when it
exists at a temperature below the boiling temperature, as it does for water at 20.0°C20.0°C. Liquid in a closed
container at a fixed temperature evaporates until the pressure of the gas reaches a certain value, called the vapor
pressure, which depends on the gas and the temperature. At this equilibrium, if heat is added, some of the liquid
will evaporate, and if heat is removed, some of the gas will condense; molecules either join the liquid or form
suspended droplets. If there is not enough liquid for the gas to reach the vapor pressure in the container, all the
liquid eventually evaporates.
If the vapor pressure of the liquid is greater than the total ambient pressure, including that of any air (or other gas),
the liquid evaporates rapidly; in other words, it boils. Thus, the boiling point of a liquid at a given pressure is the
temperature at which its vapor pressure equals the ambient pressure. Liquid and gas phases are in equilibrium at
the boiling temperature (Figure 1.14). If a substance is in a closed container at the boiling point, then the liquid is
boiling and the gas is condensing at the same rate without net change in their amounts.
Figure 1.14 Equilibrium between liquid and gas at two different boiling points inside a closed container. (a) The
rates of boiling and condensation are equal at this combination of temperature and pressure, so the liquid and gas
phases are in equilibrium. (b) At a higher temperature, the boiling rate is faster, that is, the rate at which molecules
leave the liquid and enter the gas is faster. This increases the number of molecules in the gas, which increases the
gas pressure, which in turn increases the rate at which gas molecules condense and enter the liquid. The pressure
stops increasing when it reaches the point where the boiling rate and the condensation rate are equal. The gas and
liquid are in equilibrium again at this higher temperature and pressure.
For water, 100°C100°C is the boiling point at 1.00 atm, so water and steam should exist in equilibrium under
these conditions. Why does an open pot of water at 100°C100°C boil completely away? The gas surrounding an
open pot is not pure water: it is mixed with air. If pure water and steam are in a closed container
at 100°C100°C and 1.00 atm, they will coexist—but with air over the pot, there are fewer water molecules to
condense, and water boils away. Another way to see this is that at the boiling point, the vapor pressure equals the
ambient pressure. However, part of the ambient pressure is due to air, so the pressure of the steam is less than the
vapor pressure at that temperature, and evaporation continues. Incidentally, the equilibrium vapor pressure of solids
is not zero, a fact that accounts for sublimation.
CHECK YOUR UNDERSTANDING 1.4
Check Your Understanding Explain why a cup of water (or soda) with ice cubes stays at 0°C,0°C, even on a hot
summer day.
Phase Change and Latent Heat
So far, we have discussed heat transfers that cause temperature change. However, in a phase transition, heat
transfer does not cause any temperature change.
For an example of phase changes, consider the addition of heat to a sample of ice at −20°C−20°C (Figure 1.15)
and atmospheric pressure. The temperature of the ice rises linearly, absorbing heat at a constant rate
of 2090J/kg⋅ºC2090J/kg·ºC until it reaches 0°C.0°C. Once at this temperature, the ice begins to melt and
continues until it has all melted, absorbing 333 kJ/kg of heat. The temperature remains constant at 0°C0°C during
this phase change. Once all the ice has melted, the temperature of the liquid water rises, absorbing heat at a new
constant rate of 4186 J/kg⋅ºC.4186 J/kg·ºC. At 100°C,100°C, the water begins to boil. The temperature
again remains constant during this phase change while the water absorbs 2256 kJ/kg of heat and turns into steam.
When all the liquid has become steam, the temperature rises again, absorbing heat at a rate
of 2020J/kg⋅ºC2020J/kg·ºC. If we started with steam and cooled it to make it condense into liquid water and
freeze into ice, the process would exactly reverse, with the temperature again constant during each phase transition.
Figure 1.15 Temperature versus heat. The system is constructed so that no vapor evaporates while ice warms to
become liquid water, and so that, when vaporization occurs, the vapor remains in the system. The long stretches of
constant temperatures at 0°C0°C and 100°C100°C reflect the large amounts of heat needed to cause melting and
vaporization, respectively.
Where does the heat added during melting or boiling go, considering that the temperature does not change until the
transition is complete? Energy is required to melt a solid, because the attractive forces between the molecules in the
solid must be broken apart, so that in the liquid, the molecules can move around at comparable kinetic energies;
thus, there is no rise in temperature. Energy is needed to vaporize a liquid for similar reasons. Conversely, work is
done by attractive forces when molecules are brought together during freezing and condensation. That energy must
be transferred out of the system, usually in the form of heat, to allow the molecules to stay together (Figure 1.18).
Thus, condensation occurs in association with cold objects—the glass in Figure 1.16, for example.
Figure 1.16 Condensation forms on this glass of iced tea because the temperature of the nearby air is reduced. The
air cannot hold as much water as it did at room temperature, so water condenses. Energy is released when the
water condenses, speeding the melting of the ice in the glass. (credit: Jenny Downing)
The energy released when a liquid freezes is used by orange growers when the temperature approaches 0°C0°C.
Growers spray water on the trees so that the water freezes and heat is released to the growing oranges. This
prevents the temperature inside the orange from dropping below freezing, which would damage the fruit (Figure
1.17).
Figure 1.17 The ice on these trees released large amounts of energy when it froze, helping to prevent the
temperature of the trees from dropping below 0°C0°C. Water is intentionally sprayed on orchards to help prevent
hard frosts. (credit: Hermann Hammer)
The energy involved in a phase change depends on the number of bonds or force pairs and their strength. The
number of bonds is proportional to the number of molecules and thus to the mass of the sample. The energy per
unit mass required to change a substance from the solid phase to the liquid phase, or released when the substance
changes from liquid to solid, is known as the heat of fusion. The energy per unit mass required to change a
substance from the liquid phase to the vapor phase is known as the heat of vaporization. The strength of the
forces depends on the type of molecules. The heat Q absorbed or released in a phase change in a sample of
mass m is given by
Q=mLf(melting/freezing)Q=mLf(melting/freezing)
1.7
Q=mLv(vaporization/condensation)Q=mLv(vaporization/condensation)
1.8
where the latent heat of fusion LfLf and latent heat of vaporization LvLv are material constants that are determined
experimentally. (Latent heats are also called latent heat coefficients and heats of transformation.) These constants
are “latent,” or hidden, because in phase changes, energy enters or leaves a system without causing a temperature
change in the system, so in effect, the energy is hidden.
Figure 1.18 (a) Energy is required to partially overcome the attractive forces (modeled as springs) between
molecules in a solid to form a liquid. That same energy must be removed from the liquid for freezing to take place.
(b) Molecules become separated by large distances when going from liquid to vapor, requiring significant energy to
completely overcome molecular attraction. The same energy must be removed from the vapor for condensation to
take place.
Table 1.4 lists representative values of LfLf and LvLv in kJ/kg, together with melting and boiling points. Note that in
general, Lv>LfLv>Lf. The table shows that the amounts of energy involved in phase changes can easily be
comparable to or greater than those involved in temperature changes, as Figure 1.15 and the accompanying
discussion also showed.
LfLf
Substance
Helium[2]
Hydrogen
L
Melting Point (°C)(°C)
kJ/kg
kcal/kg
Boiling Point (°C)(°C)
−272.2 (0.95 K)−272.2 (0.95 K)
5.23
1.25
−268.9(4.2K)−268.9(4.2K)
20.9
−259.3(13.9K)−259.3(13.9K)
58.6
14.0
−252.9(20.2K)−252.9(20.2K)
452
kJ/kg
LfLf
L
Nitrogen
−210.0(63.2K)−210.0(63.2K)
25.5
6.09
−195.8(77.4K)−195.8(77.4K)
201
Oxygen
−218.8(54.4K)−218.8(54.4K)
13.8
3.30
−183.0(90.2K)−183.0(90.2K)
213
Ethanol
–114
104
24.9
78.3
854
–75
332
79.3
–33.4
1370
–38.9
11.8
2.82
357
272
Water
0.00
334
79.8
100.0
2256[3]
Sulfur
119
38.1
9.10
444.6
326
Lead
327
24.5
5.85
1750
871
Antimony
631
165
39.4
1440
561
Aluminum
660
380
90
2450
11400
Silver
961
88.3
21.1
2193
2336
Gold
1063
64.5
15.4
2660
1578
Copper
1083
134
32.0
2595
5069
Uranium
1133
84
20
3900
1900
Tungsten
3410
184
44
5900
4810
Ammonia
Mercury
Table 1.4 Heats of Fusion and Vaporization[1] [1]Values quoted at the normal melting and boiling temperatures at standard
atmospheric pressure (1atm1atm). [2]Helium has no solid phase at atmospheric pressure. The melting point given is at a
pressure of 2.5 MPa. [3]At 37.0°C37.0°C (body temperature), the heat of vaporization
LvLv for water is 2430 kJ/kg or 580
kcal/kg. [4]At 37.0°C37.0°C (body temperature), the heat of vaporization, LvLv for water is 2430 kJ/kg or 580 kcal/kg.
Phase changes can have a strong stabilizing effect on temperatures that are not near the melting and boiling points,
since evaporation and condensation occur even at temperatures below the boiling point. For example, air
temperatures in humid climates rarely go above approximately 38.0°C38.0°C because most heat transfer goes
into evaporating water into the air. Similarly, temperatures in humid weather rarely fall below the dew point—the
temperature where condensation occurs given the concentration of water vapor in the air—because so much heat is
released when water vapor condenses.
More energy is required to evaporate water below the boiling point than at the boiling point, because the kinetic
energy of water molecules at temperatures below 100°C100°C is less than that at 100°C100°C, so less energy
is available from random thermal motions. For example, at body temperature, evaporation of sweat from the skin
requires a heat input of 2428 kJ/kg, which is about 10% higher than the latent heat of vaporization
at 100°C100°C. This heat comes from the skin, and this evaporative cooling effect of sweating helps reduce the
body temperature in hot weather. However, high humidity inhibits evaporation, so that body temperature might rise,
while unevaporated sweat might be left on your brow.
EXAMPLE 1.9
Calculating Final Temperature from Phase Change
Three ice cubes are used to chill a soda at 20°C20°C with mass msoda=0.25kgmsoda=0.25kg. The ice is
at 0°C0°C and each ice cube has a mass of 6.0 g. Assume that the soda is kept in a foam container so that heat
loss can be ignored and that the soda has the same specific heat as water. Find the final temperature when all ice
has melted.
Strategy
The ice cubes are at the melting temperature of 0°C.0°C. Heat is transferred from the soda to the ice for melting.
Melting yields water at 0°C,0°C, so more heat is transferred from the soda to this water until the water plus soda
system reaches thermal equilibrium.
The heat transferred to the ice is
Qice=miceLf+micecW(Tf−0°C).Qice=miceLf+micecW(Tf−0°C).
The heat given off by the soda is
Qsoda=msodacW(Tf−20°C).Qsoda=msodacW(Tf−20°C).
Since no heat is lost, Qice=−Qsoda,Qice=−Qsoda, as in Example 1.7, so that
miceLf+micecW(Tf−0°C)=−msodacW(Tf−20°C).miceLf+micecW(Tf−0°C)=−msodacW(Tf−20°C).
Solve for the unknown quantity TfTf:
Tf=msodacw(20°C)−miceLf(msoda+mice)cw.Tf=msodacw(20°C)−miceLf(msoda+mice)cw.
Solution
First we identify the known quantities. The mass of ice is mice=3×6.0g=0.018kgmice=3×6.0g=0.018kg and
the mass of soda is msoda=0.25kg.msoda=0.25kg. Then we calculate the final temperature:
Tf=20,930J−6012J1122J/°C=13°C.Tf=20,930J−6012J1122J/°C=13°C.
Significance
This example illustrates the large energies involved during a phase change. The mass of ice is about 7% of the
mass of the soda but leads to a noticeable change in the temperature of the soda. Although we assumed that the
ice was at the freezing temperature, this is unrealistic for ice straight out of a freezer: The typical temperature
is −6°C−6°C. However, this correction makes no significant change from the result we found. Can you explain
why?
Like solid-liquid and liquid-vapor transitions, direct solid-vapor transitions or sublimations involve heat. The energy
transferred is given by the equation Q=mLsQ=mLs, where LsLs is the heat of sublimation, analogous
to LfLf and LvLv. The heat of sublimation at a given temperature is equal to the heat of fusion plus the heat of
vaporization at that temperature.
We can now calculate any number of effects related to temperature and phase change. In each case, it is necessary
to identify which temperature and phase changes are taking place. Keep in mind that heat transfer and work can
cause both temperature and phase changes.
PROBLEM-SOLVING STRATEGY
The Effects of Heat Transfer
1. Examine the situation to determine that there is a change in the temperature or phase. Is there heat transfer
into or out of the system? When it is not obvious whether a phase change occurs or not, you may wish to
first solve the problem as if there were no phase changes, and examine the temperature change obtained. If
it is sufficient to take you past a boiling or melting point, you should then go back and do the problem in
steps—temperature change, phase change, subsequent temperature change, and so on.
2. Identify and list all objects that change temperature or phase.
3. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful.
4. Make a list of what is given or what can be inferred from the problem as stated (identify the knowns). If there
is a temperature change, the transferred heat depends on the specific heat of the substance (Heat Transfer,
Specific Heat, and Calorimetry), and if there is a phase change, the transferred heat depends on the latent
heat of the substance (Table 1.4).
5. Solve the appropriate equation for the quantity to be determined (the unknown).
6. Substitute the knowns along with their units into the appropriate equation and obtain numerical solutions
complete with units. You may need to do this in steps if there is more than one state to the process, such as
a temperature change followed by a phase change. However, in a calorimetry problem, each step
corresponds to a term in the single equation Qhot+Qcold=0Qhot+Qcold=0.
7. Check the answer to see if it is reasonable. Does it make sense? As an example, be certain that any
temperature change does not also cause a phase change that you have not taken into account.
Learning Objectives
By the end of this section, you will be able to:



Explain some phenomena that involve conductive, convective, and radiative heat transfer
Solve problems on the relationships between heat transfer, time, and rate of heat transfer
Solve problems using the formulas for conduction and radiation
Just as interesting as the effects of heat transfer on a system are the methods by which it occurs. Whenever there is
a temperature difference, heat transfer occurs. It may occur rapidly, as through a cooking pan, or slowly, as through
the walls of a picnic ice chest. So many processes involve heat transfer that it is hard to imagine a situation where
no heat transfer occurs. Yet every heat transfer takes place by only three methods:
1. Conduction is heat transfer through stationary matter by physical contact. (The matter is stationary on a
macroscopic scale—we know that thermal motion of the atoms and molecules occurs at any temperature
above absolute zero.) Heat transferred from the burner of a stove through the bottom of a pan to food in the
pan is transferred by conduction.
2. Convection is the heat transfer by the macroscopic movement of a fluid. This type of transfer takes place in
a forced-air furnace and in weather systems, for example.
3. Heat transfer by radiation occurs when microwaves, infrared radiation, visible light, or another form of
electromagnetic radiation is emitted or absorbed. An obvious example is the warming of Earth by the Sun. A
less obvious example is thermal radiation from the human body.
In the illustration at the beginning of this chapter, the fire warms the snowshoers’ faces largely by radiation.
Convection carries some heat to them, but most of the air flow from the fire is upward (creating the familiar shape of
flames), carrying heat to the food being cooked and into the sky. The snowshoers wear clothes designed with low
conductivity to prevent heat flow out of their bodies.
In this section, we examine these methods in some detail. Each method has unique and interesting characteristics,
but all three have two things in common: They transfer heat solely because of a temperature difference, and the
greater the temperature difference, the faster the heat transfer (Figure 1.19).
Figure 1.19 In a fireplace, heat transfer occurs by all three methods: conduction, convection, and radiation.
Radiation is responsible for most of the heat transferred into the room. Heat transfer also occurs through conduction
into the room, but much slower. Heat transfer by convection also occurs through cold air entering the room around
windows and hot air leaving the room by rising up the chimney.
CHECK YOUR UNDERSTANDING 1.6
Check Your Understanding Name an example from daily life (different from the text) for each mechanism of heat
transfer.
Conduction
As you walk barefoot across the living room carpet in a cold house and then step onto the kitchen tile floor, your feet
feel colder on the tile. This result is intriguing, since the carpet and tile floor are both at the same temperature. The
different sensation is explained by the different rates of heat transfer: The heat loss is faster for skin in contact with
the tiles than with the carpet, so the sensation of cold is more intense.
Some materials conduct thermal energy faster than others. Figure 1.20 shows a material that conducts heat
slowly—it is a good thermal insulator, or poor heat conductor—used to reduce heat flow into and out of a house.
Figure 1.20 Insulation is used to limit the conduction of heat from the inside to the outside (in winter) and from the
outside to the inside (in summer). (credit: Giles Douglas)
A molecular picture of heat conduction will help justify the equation that describes it. Figure 1.21 shows molecules in
two bodies at different temperatures, ThTh and Tc,Tc, for “hot” and “cold.” The average kinetic energy of a
molecule in the hot body is higher than in the colder body. If two molecules collide, energy transfers from the highenergy to the low-energy molecule. In a metal, the picture would also include free valence electrons colliding with
each other and with atoms, likewise transferring energy. The cumulative effect of all collisions is a net flux of heat
from the hotter body to the colder body. Thus, the rate of heat transfer increases with increasing temperature
difference ΔT=Th−Tc.ΔT=Th−Tc. If the temperatures are the same, the net heat transfer rate is zero. Because
the number of collisions increases with increasing area, heat conduction is proportional to the cross-sectional area—
a second factor in the equation.
Figure 1.21 Molecules in two bodies at different temperatures have different average kinetic energies. Collisions
occurring at the contact surface tend to transfer energy from high-temperature regions to low-temperature regions.
In this illustration, a molecule in the lower-temperature region (right side) has low energy before collision, but its
energy increases after colliding with a high-energy molecule at the contact surface. In contrast, a molecule in the
higher-temperature region (left side) has high energy before collision, but its energy decreases after colliding with a
low-energy molecule at the contact surface.
A third quantity that affects the conduction rate is the thickness of the material through which heat transfers. Figure
1.22 shows a slab of material with a higher temperature on the left than on the right. Heat transfers from the left to
the right by a series of molecular collisions. The greater the distance between hot and cold, the more time the
material takes to transfer the same amount of heat.
Figure 1.22 Heat conduction occurs through any material, represented here by a rectangular bar, whether window
glass or walrus blubber.
All four of these quantities appear in a simple equation deduced from and confirmed by experiments. The rate of
conductive heat transfer through a slab of material, such as the one in Figure 1.22, is given by
P=dQdt=kA(Th−Tc)dP=dQdt=kA(Th−Tc)d
1.9
where P is the power or rate of heat transfer in watts or in kilocalories per second, A and d are its surface area and
thickness, as shown in Figure 1.22, Th−TcTh−Tc is the temperature difference across the slab, and k is
the thermal conductivity of the material. Table 1.5 gives representative values of thermal conductivity.
More generally, we can write
P=−kAdTdx,P=−kAdTdx,
where x is the coordinate in the direction of heat flow. Since in Figure 1.22, the power and area are
constant, dT/dx is constant, and the temperature decreases linearly from ThTh to Tc.Tc.
Substance
Thermal Conductivity k (W/m⋅°C)(W/m·°C)
Diamond
2000
Silver
420
Copper
390
Gold
318
Aluminum
220
Steel iron
80
Steel (stainless)
14
Ice
2.2
Glass (average)
0.84
Concrete brick
0.84
Water
0.6
Fatty tissue (without blood)
0.2
Asbestos
0.16
Plasterboard
0.16
Wood
0.08–0.16
Snow (dry)
0.10
Cork
0.042
Glass wool
0.042
Wool
0.04
Down feathers
0.025
Thermal Conductivity k (W/m⋅°C)(W/m·°C)
Substance
Air
0.023
Polystyrene foam
0.010
Table 1.5 Thermal Conductivities of Common Substances Values are given for temperatures near 0°C0°C.
EXAMPLE 1.10
Calculating Heat Transfer through Conduction
A polystyrene foam icebox has a total area of 0.950m20.950m2 and walls with an average thickness of 2.50 cm.
The box contains ice, water, and canned beverages at 0°C.0°C. The inside of the box is kept cold by melting ice.
How much ice melts in one day if the icebox is kept in the trunk of a car at 35.0ºC35.0ºC?
Strategy
This question involves both heat for a phase change (melting of ice) and the transfer of heat by conduction. To find
the amount of ice melted, we must find the net heat transferred. This value can be obtained by calculating the rate of
heat transfer by conduction and multiplying by time.
Solution
First we identify the knowns.
k=0.010W/m⋅°Ck=0.010W/m·°C for polystyrene
foam; A=0.950m2;A=0.950m2; d=2.50cm=0.0250m;d=2.50cm=0.0250m;;Tc=0°C;Tc=0°C; Th=3
5.0°CTh=35.0°C; t=1day=24hours-86,400s.t=1day=24hours-86,400s.
Then we identify the unknowns. We need to solve for the mass of the ice, m. We also need to solve for the net heat
transferred to melt the ice, Q. The rate of heat transfer by conduction is given by
P=dQdt=kA(Th−Tc)d.P=dQdt=kA(Th−Tc)d.
The heat used to melt the ice is Q=mLfQ=mLf.We insert the known values:
P=(0.010W/m⋅°C)(0.950m2)(35.0°C−0°C)0.0250m=13.3W.P=(0.010W/m·°C)(0.950m2)(35.0°C
−0°C)0.0250m=13.3W.
Multiplying the rate of heat transfer by the time we obtain
Q=Pt=(13.3W)(86.400s)=1.15×106J.Q=Pt=(13.3W)(86.400s)=1.15×106J.
We set this equal to the heat transferred to melt the ice, Q=mLf,Q=mLf, and solve for the mass m:
m=QLf=1.15×106J334×103J/kg=3.44kg.m=QLf=1.15×106J334×103J/kg=3.44kg.
Significance
The result of 3.44 kg, or about 7.6 lb, seems about right, based on experience. You might expect to use about a 4
kg (7–10 lb) bag of ice per day. A little extra ice is required if you add any warm food or beverages.
Table 1.5 shows that polystyrene foam is a very poor conductor and thus a good insulator. Other good insulators
include fiberglass, wool, and goosedown feathers. Like polystyrene foam, these all contain many small pockets of
air, taking advantage of air’s poor thermal conductivity.
In developing insulation, the smaller the conductivity k and the larger the thickness d, the better. Thus, the ratio d/k,
called the R factor, is large for a good insulator. The rate of conductive heat transfer is inversely proportional
to R. R factors are most commonly quoted for household insulation, refrigerators, and the like. Unfortunately, in the
United States, R is still in non-metric units of ft2⋅°F⋅h/Btuft2·°F·h/Btu, although the unit usually goes unstated
[1 British thermal unit (Btu) is the amount of energy needed to change the temperature of 1.0 lb of water
by 1.0°F1.0°F, which is 1055.1 J]. A couple of representative values are an R factor of 11 for 3.5-inch-thick
fiberglass batts (pieces) of insulation and an R factor of 19 for 6.5-inch-thick fiberglass batts (Figure 1.23). In the
US, walls are usually insulated with 3.5-inch batts, whereas ceilings are usually insulated with 6.5-inch batts. In cold
climates, thicker batts may be used.
Figure 1.23 The fiberglass batt is used for insulation of walls and ceilings to prevent heat transfer between the
inside of the building and the outside environment. (credit: Tracey Nicholls)
Note that in Table 1.5, most of the best thermal conductors—silver, copper, gold, and aluminum—are also the best
electrical conductors, because they contain many free electrons that can transport thermal energy. (Diamond, an
electrical insulator, conducts heat by atomic vibrations.) Cooking utensils are typically made from good conductors,
but the handles of those used on the stove are made from good insulators (bad conductors).
EXAMPLE 1.11
Two Conductors End to End
A steel rod and an aluminum rod, each of diameter 1.00 cm and length 25.0 cm, are welded end to end. One end of
the steel rod is placed in a large tank of boiling water at 100°C100°C, while the far end of the aluminum rod is
placed in a large tank of water at 20°C20°C. The rods are insulated so that no heat escapes from their surfaces.
What is the temperature at the joint, and what is the rate of heat conduction through this composite rod?
Strategy
The heat that enters the steel rod from the boiling water has no place to go but through the steel rod, then through
the aluminum rod, to the cold water. Therefore, we can equate the rate of conduction through the steel to the rate of
conduction through the aluminum.
We repeat the calculation with a second method, in which we use the thermal resistance R of the rod, since it simply
adds when two rods are joined end to end. (We will use a similar method in the chapter on direct-current circuits.)
Solution
1. Identify the knowns and convert them to SI units.
The length of each rod is LA1=Lsteel=0.25m,LA1=Lsteel=0.25m, the cross-sectional area of each rod
is AA1=Asteel=7.85×10−5m2,AA1=Asteel=7.85×10−5m2, the thermal conductivity of aluminum
is kA1=220W/m⋅°CkA1=220W/m·°C, the thermal conductivity of steel
is ksteel=80W/m⋅°Cksteel=80W/m·°C, the temperature at the hot end is T=100°CT=100°C, and the
temperature at the cold end is T=20°CT=20°C.
2. Calculate the heat-conduction rate through the steel rod and the heat-conduction rate through the aluminum
rod in terms of the unknown temperature T at the joint:
Psteel=ksteelAsteelΔTsteelLsteel=(80W/m⋅°C)(7.85×10−5m2)(100°C−T)0.25m=(0.0251W/°C)(100°C−T);Pst
eel=ksteelAsteelΔTsteelLsteel=(80W/m·°C)(7.85×10−5m2)(100°C−T)0.25m=(0.0251W/°C)(100°C−T);
PA1=kAlAA1ΔTAlLA1=(220W/m⋅°C)(7.85×10−5m2)(T−20°C)0.25m=(0.0691W/°C)(T−20°C).PA1=kAlAA1
ΔTAlLA1=(220W/m·°C)(7.85×10−5m2)(T−20°C)0.25m=(0.0691W/°C)(T−20°C).
3. Set the two rates equal and solve for the unknown temperature:
(0.0691W/°C)(T−20°C)T==(0.0251W/°C)(100°C−T)41.3°C.(0.0691W/°C)(T−20°C)=(0.
0251W/°C)(100°C−T)T=41.3°C.
4. Calculate either rate:
Psteel=(0.0251W/°C)(100°C−41.3°C)=1.47W.Psteel=(0.0251W/°C)(100°C−41.3°C)=1.47W.
5. If desired, check your answer by calculating the other rate.
Solution
1. Recall that R=L/kR=L/k. Now P=AΔT/R,orΔT=PR/A.P=AΔT/R,orΔT=PR/A.
2. We know that ΔTsteel+ΔTAl=100°C−20°C=80°CΔTsteel+ΔTAl=100°C−20°C=80°C. We also know
that Psteel=PAl,Psteel=PAl, and we denote that rate of heat flow by P. Combine the equations:
PRsteelA+PRAlA=80°C.PRsteelA+PRAlA=80°C.
Thus, we can simply add R factors. Now, P=80°CA(Rsteel+RAl)P=80°CA(Rsteel+RAl).
3. Find the RsRs from the known quantities:
Rsteel=3.13×10−3m2⋅°C/WRsteel=3.13×10−3m2·°C/W
and
RAl=1.14×10−3m2⋅°C/W.RAl=1.14×10−3m2·°C/W.
4. Substitute these values in to find P=1.47WP=1.47W as before.
5. Determine ΔTΔT for the aluminum rod (or for the steel rod) and use it to find T at the joint.
ΔTAl=PRAlA=(1.47W)(1.14×10−3m2⋅°C/W)7.85×10−5m2=21.3°C,ΔTAl=PRAlA=(1.47W)(
1.14×10−3m2·°C/W)7.85×10−5m2=21.3°C,
so T=20°C+21.3°C=41.3°CT=20°C+21.3°C=41.3°C, as in Solution 11.
6. If desired, check by determining ΔTΔT for the other rod.
Significance
In practice, adding R values is common, as in calculating the R value of an insulated wall. In the analogous situation
in electronics, the resistance corresponds to AR in this problem and is additive even when the areas are unequal, as
is common in electronics. Our equation for heat conduction can be used only when the areas are equal; otherwise,
we would have a problem in three-dimensional heat flow, which is beyond our scope.
CHECK YOUR UNDERSTANDING 1.7
Check Your Understanding How does the rate of heat transfer by conduction change when all spatial dimensions
are doubled?
Conduction is caused by the random motion of atoms and molecules. As such, it is an ineffective mechanism for
heat transport over macroscopic distances and short times. For example, the temperature on Earth would be
unbearably cold during the night and extremely hot during the day if heat transport in the atmosphere were only
through conduction. Also, car engines would overheat unless there was a more efficient way to remove excess heat
from the pistons. The next module discusses the important heat-transfer mechanism in such situations.
Convection
In convection, thermal energy is carried by the large-scale flow of matter. It can be divided into two types. In forced
convection, the flow is driven by fans, pumps, and the like. A simple example is a fan that blows air past you in hot
surroundings and cools you by replacing the air heated by your body with cooler air. A more complicated example is
the cooling system of a typical car, in which a pump moves coolant through the radiator and engine to cool the
engine and a fan blows air to cool the radiator.
In free or natural convection, the flow is driven by buoyant forces: hot fluid rises and cold fluid sinks because density
decreases as temperature increases. The house in Figure 1.24 is kept warm by natural convection, as is the pot of
water on the stove in Figure 1.25. Ocean currents and large-scale atmospheric circulation, which result from the
buoyancy of warm air and water, transfer hot air from the tropics toward the poles and cold air from the poles toward
the tropics. (Earth’s rotation interacts with those flows, causing the observed eastward flow of air in the temperate
zones.)
Figure 1.24 Air heated by a so-called gravity furnace expands and rises, forming a convective loop that transfers
energy to other parts of the room. As the air is cooled at the ceiling and outside walls, it contracts, eventually
becoming denser than room air and sinking to the floor. A properly designed heating system using natural
convection, like this one, can heat a home quite efficiently.
Figure 1.25 Natural convection plays an important role in heat transfer inside this pot of water. Once conducted to
the inside, heat transfer to other parts of the pot is mostly by convection. The hotter water expands, decreases in
density, and rises to transfer heat to other regions of the water, while colder water sinks to the bottom. This process
keeps repeating.
INTERACTIVE
Natural convection like that of Figure 1.24 and Figure 1.25, but acting on rock in Earth’s mantle, drives plate
tectonics that are the motions that have shaped Earth’s surface.
Convection is usually more complicated than conduction. Beyond noting that the convection rate is often
approximately proportional to the temperature difference, we will not do any quantitative work comparable to the
formula for conduction. However, we can describe convection qualitatively and relate convection rates to heat and
time. Air is a poor conductor, so convection dominates heat transfer by air. Therefore, the amount of available space
for airflow determines whether air transfers heat rapidly or slowly. There is little heat transfer in a space filled with air
with a small amount of other material that prevents flow. The space between the inside and outside walls of a typical
American house, for example, is about 9 cm (3.5 in.)—large enough for convection to work effectively. The addition
of wall insulation prevents airflow, so heat loss (or gain) is decreased. On the other hand, the gap between the two
panes of a double-paned window is about 1 cm, which largely prevents convection and takes advantage of air’s low
conductivity reduce heat loss. Fur, cloth, and fiberglass also take advantage of the low conductivity of air by trapping
it in spaces too small to support convection (Figure 1.26).
Figure 1.26 Fur is filled with air, breaking it up into many small pockets. Convection is very slow here, because the
loops are so small. The low conductivity of air makes fur a very good lightweight insulator.
Some interesting phenomena happen when convection is accompanied by a phase change. The combination allows
us to cool off by sweating even if the temperature of the surrounding air exceeds body temperature. Heat from the
skin is required for sweat to evaporate from the skin, but without air flow, the air becomes saturated and evaporation
stops. Air flow caused by convection replaces the saturated air by dry air and evaporation continues.
EXAMPLE 1.12
Calculating the Flow of Mass during Convection
The average person produces heat at the rate of about 120 W when at rest. At what rate must water evaporate from
the body to get rid of all this energy? (For simplicity, we assume this evaporation occurs when a person is sitting in
the shade and surrounding temperatures are the same as skin temperature, eliminating heat transfer by other
methods.)
Strategy
Energy is needed for this phase change (Q=mLvQ=mLv). Thus, the energy loss per unit time is
Qt=mLVt=120W=120 J/s.Qt=mLVt=120W=120 J/s.
We divide both sides of the equation by LvLv to find that the mass evaporated per unit time is
mt=120J/sLv.mt=120J/sLv.
Solution
Insert the value of the latent heat from Table 1.4, Lv=2430kJ/kg=2430J/gLv=2430kJ/kg=2430J/g. This
yields
mt=120J/s2430J/g=0.0494g/s=2.96g/min.mt=120J/s2430J/g=0.0494g/s=2.96g/min.
Significance
Evaporating about 3 g/min seems reasonable. This would be about 180 g (about 7 oz.) per hour. If the air is very
dry, the sweat may evaporate without even being noticed. A significant amount of evaporation also takes place in
the lungs and breathing passages.
Another important example of the combination of phase change and convection occurs when water evaporates from
the oceans. Heat is removed from the ocean when water evaporates. If the water vapor condenses in liquid droplets
as clouds form, possibly far from the ocean, heat is released in the atmosphere. Thus, there is an overall transfer of
heat from the ocean to the atmosphere. This process is the driving power behind thunderheads, those great
cumulus clouds that rise as much as 20.0 km into the stratosphere (Figure 1.27). Water vapor carried in by
convection condenses, releasing tremendous amounts of energy. This energy causes the air to expand and rise to
colder altitudes. More condensation occurs in these regions, which in turn drives the cloud even higher. This
mechanism is an example of positive feedback, since the process reinforces and accelerates itself. It sometimes
produces violent storms, with lightning and hail. The same mechanism drives hurricanes.
INTERACTIVE
This time-lapse video shows convection currents in a thunderstorm, including “rolling” motion similar to that of
boiling water.
Figure 1.27 Cumulus clouds are caused by water vapor that rises because of convection. The rise of clouds is
driven by a positive feedback mechanism. (credit: “Amada44”/Wikimedia Commons)
CHECK YOUR UNDERSTANDING 1.8
Check Your Understanding Explain why using a fan in the summer feels refreshing.
Radiation
You can feel the heat transfer from the Sun. The space between Earth and the Sun is largely empty, so the Sun
warms us without any possibility of heat transfer by convection or conduction. Similarly, you can sometimes tell that
the oven is hot without touching its door or looking inside—it may just warm you as you walk by. In these examples,
heat is transferred by radiation (Figure 1.28). That is, the hot body emits electromagnetic waves that are absorbed
by the skin. No medium is required for electromagnetic waves to propagate. Different names are used for
electromagnetic waves of different wavelengths: radio waves, microwaves, infrared radiation, visible light, ultraviolet
radiation, X-rays, and gamma rays.
Figure 1.28 Most of the heat transfer from this fire to the observers occurs through infrared radiation. The visible
light, although dramatic, transfers relatively little thermal energy. Convection transfers energy away from the
observers as hot air rises, while conduction is negligibly slow here. Skin is very sensitive to infrared radiation, so you
can sense the presence of a fire without looking at it directly. (credit: Daniel O’Neil)
The energy of electromagnetic radiation varies over a wide range, depending on the wavelength: A shorter
wavelength (or higher frequency) corresponds to a higher energy. Because more heat is radiated at higher
temperatures, higher temperatures produce more intensity at every wavelength but especially at shorter
wavelengths. In visible light, wavelength determines color—red has the longest wavelength and violet the shortest—
so a temperature change is accompanied by a color change. For example, an electric heating element on a stove
glows from red to orange, while the higher-temperature steel in a blast furnace glows from yellow to white. Infrared
radiation is the predominant form radiated by objects cooler than the electric element and the steel. The radiated
energy as a function of wavelength depends on its intensity, which is represented in Figure 1.29 by the height of the
distribution. (Electromagnetic Waves explains more about the electromagnetic spectrum, and Photons and Matter
Waves discusses why the decrease in wavelength corresponds to an increase in energy.)
Figure 1.29 (a) A graph of the spectrum of electromagnetic waves emitted from an ideal radiator at three different
temperatures. The intensity or rate of radiation emission increases dramatically with temperature, and the spectrum
shifts down in wavelength toward the visible and ultraviolet parts of the spectrum. The shaded portion denotes the
visible part of the spectrum. It is apparent that the shift toward the ultraviolet with temperature makes the visible
appearance shift from red to white to blue as temperature increases. (b) Note the variations in color corresponding
to variations in flame temperature.
The rate of heat transfer by radiation also depends on the object’s color. Black is the most effective, and white is the
least effective. On a clear summer day, black asphalt in a parking lot is hotter than adjacent gray sidewalk, because
black absorbs better than gray (Figure 1.30). The reverse is also true—black radiates better than gray. Thus, on a
clear summer night, the asphalt is colder than the gray sidewalk, because black radiates the energy more rapidly
than gray. A perfectly black object would be an ideal radiator and an ideal absorber, as it would capture all the
radiation that falls on it. In contrast, a perfectly white object or a perfect mirror would reflect all radiation, and a
perfectly transparent object would transmit it all (Figure 1.31). Such objects would not emit any radiation.
Mathematically, the color is represented by the emissivity e. A “blackbody” radiator would have an e=1e=1,
whereas a perfect reflector or transmitter would have e=0e=0. For real examples, tungsten light bulb filaments
have an e of about 0.5, and carbon black (a material used in printer toner) has an emissivity of about 0.95.
Figure 1.30 The darker pavement is hotter than the lighter pavement (much more of the ice on the right has
melted), although both have been in the sunlight for the same time. The thermal conductivities of the pavements are
the same.
Figure 1.31 A black object is a good absorber and a good radiator, whereas a white, clear, or silver object is a poor
absorber and a poor radiator.
To see that, consider a silver object and a black object that can exchange heat by radiation and are in thermal
equilibrium. We know from experience that they will stay in equilibrium (the result of a principle that will be discussed
at length in Second Law of Thermodynamics). For the black object’s temperature to stay constant, it must emit as
much radiation as it absorbs, so it must be as good at radiating as absorbing. Similar considerations show that the
silver object must radiate as little as it absorbs. Thus, one property, emissivity, controls both radiation and
absorption.
Finally, the radiated heat is proportional to the object’s surface area, since every part of the surface radiates. If you
knock apart the coals of a fire, the radiation increases noticeably due to an increase in radiating surface area.
The rate of heat transfer by emitted radiation is described by the Stefan-Boltzmann law of radiation:
P=σAeT4,P=σAeT4,
where σ=5.67×10−8J/s⋅m2⋅K4σ=5.67×10−8J/s·m2·K4 is the Stefan-Boltzmann constant, a combination of
fundamental constants of nature; A is the surface area of the object; and T is its temperature in kelvins.
The proportionality to the fourth power of the absolute temperature is a remarkably strong temperature dependence.
It allows the detection of even small temperature variations. Images called thermographs can be used medically to
detect regions of abnormally high temperature in the body, perhaps indicative of disease. Similar techniques can be
used to detect heat leaks in homes (Figure 1.32), optimize performance of blast furnaces, improve comfort levels in
work environments, and even remotely map Earth’s temperature profile.
Figure 1.32 A thermograph of part of a building shows temperature variations, indicating where heat transfer to the
outside is most severe. Windows are a major region of heat transfer to the outside of homes. (credit: US Army)
The Stefan-Boltzmann equation needs only slight refinement to deal with a simple case of an object’s absorption of
radiation from its surroundings. Assuming that an object with a temperature T1T1 is surrounded by an environment
with uniform temperature T2T2, the net rate of heat transfer by radiation is
Pnet=σeA(T24−T14),Pnet=σeA(T24−T14),
1.10
where e is the emissivity of the object alone. In other words, it does not matter whether the surroundings are white,
gray, or black: The balance of radiation into and out of the object depends on how well it emits and absorbs
radiation. When T2>T1,T2>T1, the quantity PnetPnet is positive, that is, the net heat transfer is from hot to cold.
Before doing an example, we have a complication to discuss: different emissivities at different wavelengths. If the
fraction of incident radiation an object reflects is the same at all visible wavelengths, the object is gray; if the fraction
depends on the wavelength, the object has some other color. For instance, a red or reddish object reflects red light
more strongly than other visible wavelengths. Because it absorbs less red, it radiates less red when hot. Differential
reflection and absorption of wavelengths outside the visible range have no effect on what we see, but they may
have physically important effects. Skin is a very good absorber and emitter of infrared radiation, having an emissivity
of 0.97 in the infrared spectrum. Thus, in spite of the obvious variations in skin color, we are all nearly black in the
infrared. This high infrared emissivity is why we can so easily feel radiation on our skin. It is also the basis for the
effectiveness of night-vision scopes used by law enforcement and the military to detect human beings.
EXAMPLE 1.13
Calculating the Net Heat Transfer of a Person
What is the rate of heat transfer by radiation of an unclothed person standing in a dark room whose ambient
temperature is 22.0°C22.0°C? The person has a normal skin temperature of 33.0°C33.0°C and a surface area
of 1.50m2.1.50m2. The emissivity of skin is 0.97 in the infrared, the part of the spectrum where the radiation takes
place.
Strategy
We can solve this by using the equation for the rate of radiative heat transfer.
Solution
Insert the temperature values T2=295KT2=295K and T1=306KT1=306K, so that
Qt=σeA(T24−T14)=(5.67×10−8J/s⋅m2⋅K4)(0.97)(1.50m2)[(295K)4−(306K)4]=−99J/s=−99
W.Qt=σeA(T24−T14)=(5.67×10−8J/s·m2·K4)(0.97)(1.50m2)[(295K)4−(306K)4]=−99J/s=−99W.
Significance
This value is a significant rate of heat transfer to the environment (note the minus sign), considering that a person at
rest may produce energy at the rate of 125 W and that conduction and convection are also transferring energy to the
environment. Indeed, we would probably expect this person to feel cold. Clothing significantly reduces heat transfer
to the environment by all mechanisms, because clothing slows down both conduction and convection, and has a
lower emissivity (especially if it is light-colored) than skin.
The average temperature of Earth is the subject of much current discussion. Earth is in radiative contact with both
the Sun and dark space, so we cannot use the equation for an environment at a uniform temperature. Earth
receives almost all its energy from radiation of the Sun and reflects some of it back into outer space. Conversely,
dark space is very cold, about 3 K, so that Earth radiates energy into the dark sky. The rate of heat transfer from soil
and grasses can be so rapid that frost may occur on clear summer evenings, even in warm latitudes.
The average temperature of Earth is determined by its energy balance. To a first approximation, it is the
temperature at which Earth radiates heat to space as fast as it receives energy from the Sun.
An important parameter in calculating the temperature of Earth is its emissivity (e). On average, it is about 0.65, but
calculation of this value is complicated by the great day-to-day variation in the highly reflective cloud coverage.
Because clouds have lower emissivity than either oceans or land masses, they emit some of the radiation back to
the surface, greatly reducing heat transfer into dark space, just as they greatly reduce heat transfer into the
atmosphere during the day. There is negative feedback (in which a change produces an effect that opposes that
change) between clouds and heat transfer; higher temperatures evaporate more water to form more clouds, which
reflect more radiation back into space, reducing the temperature.
The often-mentioned greenhouse effect is directly related to the variation of Earth’s emissivity with wavelength
(Figure 1.33). The greenhouse effect is a natural phenomenon responsible for providing temperatures suitable for
life on Earth and for making Venus unsuitable for human life. Most of the infrared radiation emitted from Earth is
absorbed by carbon dioxide (CO2CO2) and water (H2OH2O) in the atmosphere and then re-radiated into outer
space or back to Earth. Re-radiation back to Earth maintains its surface temperature about 40°C40°C higher than
it would be if there were no atmosphere. (The glass walls and roof of a greenhouse increase the temperature inside
by blocking convective heat losses, not radiative losses.)
Figure 1.33 The greenhouse effect is the name given to the increase of Earth’s temperature due to absorption of
radiation in the atmosphere. The atmosphere is transparent to incoming visible radiation and most of the Sun’s
infrared. The Earth absorbs that energy and re-emits it. Since Earth’s temperature is much lower than the Sun’s, it
re-emits the energy at much longer wavelengths, in the infrared. The atmosphere absorbs much of that infrared
radiation and radiates about half of the energy back down, keeping Earth warmer than it would otherwise be. The
amount of trapping depends on concentrations of trace gases such as carbon dioxide, and an increase in the
concentration of these gases increases Earth’s surface temperature.
The greenhouse effect is central to the discussion of global warming due to emission of carbon dioxide and methane
(and other greenhouse gases) into Earth’s atmosphere from industry, transportation, and farming. Changes in global
climate could lead to more intense storms, precipitation changes (affecting agriculture), reduction in rain forest
biodiversity, and rising sea levels.
INTERACTIVE
You can explore a simulation of the greenhouse effect that takes the point of view that the atmosphere scatters
(redirects) infrared radiation rather than absorbing it and reradiating it. You may want to run the simulation first with
no greenhouse gases in the atmosphere and then look at how adding greenhouse gases affects the infrared
radiation from the Earth and the Earth’s temperature.
PROBLEM-SOLVING STRATEGY
Effects of Heat Transfer
1.
2.
3.
4.
5.
Examine the situation to determine what type of heat transfer is involved.
Identify the type(s) of heat transfer—conduction, convection, or radiation.
Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful.
Make a list of what is given or what can be inferred from the problem as stated (identify the knowns).
Solve the appropriate equation for the quantity to be determined (the unknown).
6. For conduction, use the equation P=kAΔTdP=kAΔTd. Table 1.5 lists thermal conductivities. For convection,
determine the amount of matter moved and the equation Q=mcΔTQ=mcΔT, along
with Q=mLfQ=mLf or Q=mLVQ=mLV if a substance changes phase. For radiation, the
equation Pnet=σeA(T24−T14)Pnet=σeA(T24−T14) gives the net heat transfer rate.
7. Substitute the knowns along with their units into the appropriate equation and obtain numerical solutions
complete with units.
8. Check the answer to see if it is reasonable. Does it make sense?
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